Problem 65 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$x^{y}=y^{x}$$

Step-by-Step Solution

Answer

The derivative \(\frac{dy}{dx}\) is \(y' = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}\)."

Step 1: Logarithmic Differentiation

Take the natural logarithm of both sides of the equation \(x^{y} = y^{x}\) to make differentiation easier:\[\ln(x^y) = \ln(y^x)\]Apply the logarithm power rule:\[y \ln(x) = x \ln(y)\]

Step 2: Differentiate Both Sides Implicitly

Differentiate both sides with respect to \(x\):\[\frac{d}{dx}[y \ln(x)] = \frac{d}{dx}[x \ln(y)]\]Apply the product rule and chain rule where necessary:\[(y' \ln(x) + \frac{y}{x}) = (\ln(y) + x \frac{y'}{y})\]

Step 3: Isolate \(y'\)

Rearrange the differentiated equation to isolate \(y'\):\[y' \ln(x) - x \frac{y'}{y} = \ln(y) - \frac{y}{x}\]Factor out \(y'\):\[y'(\ln(x) - \frac{x}{y}) = \ln(y) - \frac{y}{x}\]Finally, solve for \(y'\):\[y' = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}\]

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful tool for differentiating complex functions, especially when the variables are raised to the power of another variable. In this exercise, our equation is given as \(x^y = y^x\). Taking the natural logarithm of both sides simplifies this expression. By using the logarithm power rule, which states that \(\log(a^b) = b \log(a)\), we can transform the initial equation into \(y \ln(x) = x \ln(y)\).

This approach is particularly advantageous because it turns products and powers into sums and products, which are easier to differentiate. Logarithmic differentiation involves:

• Taking the natural logarithm of both sides to simplify the differentiation process.
• Applying the logarithmic identity \(\log(a^b) = b \log(a)\) to rewrite the expression.

Once the equation is simplified through logarithmic manipulation, we can proceed with implicit differentiation to find the derivative \(dy/dx\).

Product Rule

The product rule is essential when differentiating expressions involving products of functions. When we differentiate \(y \ln(x)\) and \(x \ln(y)\), we need the product rule to find the derivatives correctly. The product rule states: if \(u(x)\) and \(v(x)\) are functions of \(x\), then \(d(uv)/dx = u'v + uv'\).

In the original solution:

• \(\frac{d}{dx}[y \ln(x)]\) requires the application of the product rule. Here, \(u = y\) and \(v = \ln(x)\).
• \(\frac{d}{dx}[x \ln(y)]\) also uses the product rule with \(u = x\) and \(v = \ln(y)\).

By applying the product rule, we differentiate each part while holding the other constant, ensuring we accurately capture changes in each component of the product as affected by \(x\).

Chain Rule

The chain rule is another key element when differentiating complex functions, especially those involving composite functions. The chain rule states that if a function \(y = g(f(x))\) is the composition of \(f\) and \(g\), then its derivative is \(dy/dx = g'(f(x)) \cdot f'(x)\).

In our exercise, the chain rule is applied where necessary:

• When differentiating \(\ln(y)\), related to the function \(\ln(g(x))\), we use the chain rule because \(y\) itself is a function of \(x\).
• This results in a term \(\frac{y'}{y}\), indicating the derivative of \(y\) with respect to \(x\) times the derivative of \(\ln(y)\) with respect to \(y\).

Combined with implicit differentiation, the chain rule helps incorporate derivatives of nested functions effectively, ensuring a comprehensive solution to the problem.

Algebraic Manipulation

Algebraic manipulation is crucial in rearranging equations to isolate the desired variable, especially after implicit differentiation. In our problem, once both sides of the equation are differentiated, we need to isolate \(y'\) to find the derivative \(dy/dx\).

To isolate \(y'\), we:

• Reorganize the terms resulting from implicit differentiation to cluster all \(y'\) terms on one side of the equation.
• Factor out \(y'\) from the terms on its side.
• Solve for \(y'\), which involves straightforward algebra to isolate it completely.

This approach simplifies the differentiation process, ensuring accurate calculation of derivatives by handling the mathematical expressions effectively. Through algebraic manipulation, expressions are restructured to allow as simple a path as possible to the solution.